Many metals react with oxygen gas to form the metal oxide. For example calcium reacts:
2Ca(s) + O2(g)--> 2CaO(s)
Calculate the mass of calcium prepared from 4,20g of Ca and 2.80g of O2. (a) How many moles of CaO can form from the given mass of Ca? (b) How many moles of CaO can form from the given mass of O2? (c)What is the limiting reactant? (d) How many grams of CaO can be formed?
Well, let's dive into this chemistry problem, shall we? Don't worry, I promise to make it as exciting as watching water boil!
First, let's calculate the number of moles of calcium (Ca) from the given mass. To do this, we divide the mass by the molar mass of Ca, which is approximately 40.08 g/mol.
(a) Moles of Ca = mass of Ca / molar mass of Ca
= 4.20 g / 40.08 g/mol
≈ 0.105 mol
Now, let's determine how many moles of calcium oxide (CaO) can form from the given mass of Ca. Looking at the balanced chemical equation, we see that for every 2 moles of Ca, 2 moles of CaO are produced.
(b) Moles of CaO from given mass of Ca = 0.105 mol * (2 mol CaO / 2 mol Ca)
= 0.105 mol
Next up, we need to calculate the number of moles of oxygen (O2) from the given mass. To find the moles, we divide the mass by the molar mass of O2, which is about 32.00 g/mol.
(c) Moles of O2 = mass of O2 / molar mass of O2
= 2.80 g / 32.00 g/mol
≈ 0.0875 mol
Now, let's determine the moles of CaO that can form from the given mass of O2. From the balanced chemical equation, we know that 1 mole of O2 reacts with 2 moles of CaO.
(d) Moles of CaO from given mass of O2 = 0.0875 mol * (2 mol CaO / 1 mol O2)
= 0.175 mol
The limiting reactant is the one that is completely consumed in the reaction. It is the reactant that produces the lesser amount of product. In this case, we can see that the number of moles of CaO from Ca (0.105 mol) is less than the number of moles from O2 (0.175 mol). Therefore, calcium (Ca) is the limiting reactant.
Finally, to calculate the mass of CaO that can be formed, we multiply the moles of CaO from calcium by its molar mass, which is approximately 56.08 g/mol.
(e) Mass of CaO formed = moles of CaO from calcium * molar mass of CaO
= 0.105 mol * 56.08 g/mol
≈ 5.894 g
So, approximately 5.894 grams of calcium oxide (CaO) can be formed in this reaction. I hope I was able to keep you entertained while we figured this out!
To calculate the mass of calcium prepared from 4.20g of Ca and 2.80g of O2, we need to follow the steps below:
(a) Calculate the moles of Ca from the given mass of Ca:
Molar mass of Ca = 40.08 g/mol
Number of moles of Ca = Mass of Ca / Molar mass of Ca
Number of moles of Ca = 4.20g / 40.08 g/mol
(b) Calculate the moles of O2 from the given mass of O2:
Molar mass of O2 = 32.00 g/mol
Number of moles of O2 = Mass of O2 / Molar mass of O2
Number of moles of O2 = 2.80g / 32.00 g/mol
(c) Determine the limiting reactant:
To determine the limiting reactant, we compare the moles of each reactant used in the reaction. The reactant that produces the least amount of product is the limiting reactant.
From the balanced equation, we can see that the molar ratio between Ca and CaO is 2:2. Therefore, the moles of CaO formed will be equal to the moles of Ca used.
(d) Calculate the moles of CaO that can form:
Number of moles of CaO = Number of moles of Ca (since they have a 1:1 ratio)
Number of moles of CaO = 4.20g / 40.08 g/mol
Now, to calculate the mass of CaO formed, we use the molar mass of CaO:
Molar mass of CaO = 56.08 g/mol
Mass of CaO formed = Number of moles of CaO x Molar mass of CaO
Substitute the value of moles of CaO into the equation:
Mass of CaO formed = (4.20g / 40.08 g/mol) x 56.08 g/mol
Calculate the value to find the answer.
To calculate the mass of calcium prepared from 4.20g of Ca and 2.80g of O2, we need to follow a step-by-step approach:
(a) Calculate the moles of Ca from the given mass of Ca:
To do this, we will use the molar mass of calcium, which is 40.08 g/mol.
Number of moles of Ca = mass of Ca / molar mass of Ca
Number of moles of Ca = 4.20g / 40.08 g/mol
(b) Calculate the moles of O2 from the given mass of O2:
To do this, we will use the molar mass of oxygen, which is 32.00 g/mol.
Number of moles of O2 = mass of O2 / molar mass of O2
Number of moles of O2 = 2.80g / 32.00 g/mol
(c) Determine the limiting reactant:
To find the limiting reactant, we need to compare the moles of CaO that can be formed from each reactant. The reactant that produces fewer moles of CaO will be the limiting reactant.
According to the balanced equation, 2 moles of Ca react with 1 mole of O2 to form 2 moles of CaO.
From part (a), we have the moles of Ca.
From part (b), we have the moles of O2.
Using the mole ratio from the balanced equation, we can calculate the moles of CaO that can form from each reactant:
Moles of CaO from Ca = (Number of moles of Ca) x (2 moles of CaO / 2 moles of Ca)
Moles of CaO from O2 = (Number of moles of O2) x (2 moles of CaO / 1 mole of O2)
(d) Calculate the mass of CaO that can be formed:
To calculate this, we will use the molar mass of CaO, which is 56.08 g/mol.
Mass of CaO = (Moles of CaO from Ca + Moles of CaO from O2) x molar mass of CaO
Mass of CaO = (Moles of CaO from Ca + Moles of CaO from O2) x 56.08 g/mol
Following these steps should allow you to calculate the mass of calcium prepared from the given amounts of Ca and O2, as well as the other requested information.
Many metals react with oxygen gas to form the metal oxide. For example calcium reacts:
2Ca(s) + O2(g)--> 2CaO(s)
Calculate the mass of calcium prepared from 4,20g of Ca and 2.80g of O2. (a) How many moles of CaO can form from the given mass of Ca?
4.20 g Ca x (1 mole Ca/40.08 g) = ?? moles Ca. Moles CaO are same since 2 moles Ca produce 2 mols CaO.
(b) How many moles of CaO can form from the given mass of O2?
2.80 g O2 x (1 mole O2/32 g ) = ?? moles O2. Moles CaO will be twice moles O2 since 1 mole O2 produces 2 moles CaO
(c)What is the limiting reactant?
The limiting reagent is the one producing the smaller number of moles of CaO.
(d) How many grams of CaO can be formed?
Smaller number moles CaO x molar mass CaO = grams CaO formed.